Rewrite the function by completing the square. $f(x)=x^{2}+6x-78$ $f(x)=(x+$
Answer: We want to complete $x^2{+6}x$ into a perfect square. To do that, we should add $\left(\dfrac{{+6}}{2}\right)^2={9}$ to it: $x^2{+6}x+{9}=(x+3)^2$ In order to keep the expression equivalent, we add and subtract ${9}$, not forgetting the expression's constant term, $-78$ : $\begin{aligned} f(x)&=x^2+6x-78 \\\\ &=x^2+6x+{9}-78-{9} \\\\ &=(x+3)^2-78-9 \\\\ &=(x+3)^2-87 \end{aligned}$ In conclusion, after completing the square, the function is written as $f(x)=(x + 3)^2 - 87$